3.7 to 5v ?

If you have an old auxiliary backup battery for a cell phone there is a circuit inside that takes the 3.7 volts of the battery and boosts it to 5 volts. This circuit will also allow you to charge the 3.7 Lithium battery with a standard phone charger. The only catch is that it may require a certain level of output current to remain on. In other words the 5 V device that you want to power with the battery might have to draw 20 or more mA to keep the circuit engaged.

John

Again, the device calls for "5V 1A". So if it requires "1A" and I have 11.1v - 3.5a and I step it down, or I have 3.7 with 10.5a with a step up, wouldnt that work ? 

Hi OT

That is a good question. If your device is designed to run on 5 Volts and it draws 1 Amp yes you can obtain this from either a step up converter (called a boost converter) or from a step down converter (called a buck converter). I will assume 100% efficiency for these conversions. In the case of the 3.7 volt boost conversion the converter will draw 1.35 Amps from the 3.7 volt battery in order to deliver the 1 Amp at 5 Volts. If the efficiency is less that 100% AND IT ALWAYS IS the current draw from the 3.7 volt battery will be more than 1.35 Amps. In the case of the 11.1 Volts and a buck converter the converter will draw at least 0.45 Amps from the 11.1 Volt battery in order to deliver 1 Amp at 5 Volts. There are DC to Dc converters on sale on ebay that will allow you to do either of these conversions.

If you want to build a converter yourself there are some integrated circuits available that allow you to build a converter by adding the correct resistors, capacitors, and inductors. Unless you have a good back ground in electronics it might be a fairly steep learning curve as a first project.

I want to make a clarification as I get the impression from your question that you may think that the battery or power supply decides the amount of current that it will supply to the device. The battery or converter has an internal resistance and provides a voltage. The device also has a resistance. When the voltage of the battery is put on the device a current flows through the resistance of the battery and through the resistance of the device. This current is equal to the Voltage of the battery divided by the added resistance of the battery and the device and is measured in Amps. I could have a 5 volt battery as big as a house and I would get the same 1 amp current when I hooked up your device that I would get if I hooked it up to a much smaller 5 Volt battery. For this reason batteries do not come with an amperage listed on them. What you may see however is an Amp Hour or milli-Amp hour rating. These ratings give us some idea how much power is in the battery. The device that we put on the battery will determine how much current in amps flow in the circuit and if we divide that measure of Amps into the Amp Hour rating of the battery we get a rough idea of how long the battery will run the device before it runs out of power.

If I have missed your question you will have to rephrase it and try again.

John

Hi OT

That is a good question. If your device is designed to run on 5 Volts and it draws 1 Amp yes you can obtain this from either a step up converter (called a boost converter) or from a step down converter (called a buck converter). I will assume 100% efficiency for these conversions. In the case of the 3.7 volt boost conversion the converter will draw 1.35 Amps from the 3.7 volt battery in order to deliver the 1 Amp at 5 Volts. If the efficiency is less that 100% AND IT ALWAYS IS the current draw from the 3.7 volt battery will be more than 1.35 Amps. In the case of the 11.1 Volts and a buck converter the converter will draw at least 0.45 Amps from the 11.1 Volt battery in order to deliver 1 Amp at 5 Volts. There are DC to Dc converters on sale on ebay that will allow you to do either of these conversions.

If you want to build a converter yourself there are some integrated circuits available that allow you to build a converter by adding the correct resistors, capacitors, and inductors. Unless you have a good back ground in electronics it might be a fairly steep learning curve as a first project.

I want to make a clarification as I get the impression from your question that you may think that the battery or power supply decides the amount of current that it will supply to the device. The battery or converter has an internal resistance and provides a voltage. The device also has a resistance. When the voltage of the battery is put on the device a current flows through the resistance of the battery and through the resistance of the device. This current is equal to the Voltage of the battery divided by the added resistance of the battery and the device and is measured in Amps. I could have a 5 volt battery as big as a house and I would get the same 1 amp current when I hooked up your device that I would get if I hooked it up to a much smaller 5 Volt battery. For this reason batteries do not come with an amperage listed on them. What you may see however is an Amp Hour or milli-Amp hour rating. These ratings give us some idea how much power is in the battery. The device that we put on the battery will determine how much current in amps flow in the circuit and if we divide that measure of Amps into the Amp Hour rating of the battery we get a rough idea of how long the battery will run the device before it runs out of power.

If I have missed your question you will have to rephrase it and try again.

John

That is a good question. If your device is designed to run on 5 Volts and it draws 1 Amp yes you can obtain this from either a step up converter (called a boost converter) or from a step down converter (called a buck converter). I will assume 100% efficiency for these conversions. In the case of the 3.7 volt boost conversion the converter will draw 1.35 Amps from the 3.7 volt battery in order to deliver the 1 Amp at 5 Volts. If the efficiency is less that 100% AND IT ALWAYS IS the current draw from the 3.7 volt battery will be more than 1.35 Amps. In the case of the 11.1 Volts and a buck converter the converter will draw at least 0.45 Amps from the 11.1 Volt battery in order to deliver 1 Amp at 5 Volts. There are DC to Dc converters on sale on ebay that will allow you to do either of these conversions.

Hello John,

So would you say it would be better to go with a series circuit configuration for more efficiency and less heat ?

Oscar

Hi Oscar,

As long as you use a switch mode converter, whether the batteries are in series or parallel  it will both work about the same for heat loss. If you are going to try to charge the batteries in the unit the Parallel would have some advantage as mentioned by a previous poster. Otherwise it is your choice and may also depend on whether you find a buck or a boost converter that you like.

John